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3 years ago

GIVING OUT BRAINLIEST !!! HELP ME PLSS

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

6 0

Answer:

Step-by-step explanation:

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Need help (easy 20 points)

schepotkina [342]

Answer: well I Might be wrong but all I have to do is look it up.

Step-by-step explanation:

4 0

2 years ago

Math question thanks!

skelet666 [1.2K]

Answer:

B C and E

Step-by-step explanation:

The easiest way to know if the graph was a function is the vertical line test

If you can draw a verticle line that interests the graph two or more times it is not a function and if you can't it is a function

4 0

3 years ago

Surface Area of Cylinder=2πrh+2πr2

Alex

Find the surface area when r is 8 inches and h is 8 inches.

$\qquad$ A. 160π in²

$\qquad$ B. 154π in²

$\qquad$ C. 288π in²

$\qquad$ D. 256π in² ☑

We are given –

$\qquad$ ⇢Radius of cylinder , r = 8 inches

$\qquad$ ⇢ Height of cylinder, h = 8 inches.

We are asked to find surface area of the given cylinder.

Formula to find the surface cylinder given by –

$\bf \star \pink{ Surface\: Area_{(Cylinder)} = 2\pi rh +2\pi r^2 }$

Now, Substitute given values –

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2\pi rh +2\pi r^2$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2 \pi \times 8 \times 8 + 2\pi \times 8^2$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2 \pi \times 8^2 + 2\pi \times 8^2$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2\pi \times 64 +2\pi \times 64$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} =128 \pi + 128\pi$

$\purple{\bf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 256\pi \: in^2 }$

Henceforth,Option D is the correct answer.

8 0

2 years ago

Find the x-intercepts of the parabola with vertex (1,-108) and y-intercept (0,-105). Write your answer in this form: (X1,y1), (x

Stolb23 [73]

Answer:

(7, 0) and (-5, 0)

Step-by-step explanation:

<u>Vertex form</u>

$y=a(x-h)^2+k$

(where (h, k) is the vertex)

Given:

vertex = (1, -108)

$\implies y=a(x-1)^2-108$

Given:

y-intercept = (0, -105)

$\implies a(0-1)^2-108=-105$

$\implies a(-1)^2=-105+108$

$\implies a=3$

Therefore:

$\implies y=3(x-1)^2-108$

The x-intercepts are when y = 0

$\implies 3(x-1)^2-108=0$

$\implies 3(x-1)^2=108$

$\implies (x-1)^2=36$

$\implies x-1=\pm \sqrt{36}$

$\implies x=1\pm 6$

$\implies x=7, x=-5$

Therefore, the x-intercepts are (7, 0) and (-5, 0)

7 0

2 years ago

I need help on this

Alisiya [41]

2. rhombus

3. rectangle

4. parallelogram

5.trapezoid

6.quadrilateral

6 0

3 years ago

Read 2 more answers